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Effectus [21]
3 years ago
10

A scientist plants two rows of corn for experimentation. She puts fertilizer on row 1 but does not put fertilizer on row 2. Both

rows receive the same amount of sun and water. She checks the growth of the corn over the course of five months.
Q9) What is the independent variable in this experiment? Why is this variable independent?


Q10) What is the dependent variable in this experiment? Why is this a dependent variable?

Q11) What variables are controlled in this experiment? Why are/is these/this variable control variable(s)?
Chemistry
2 answers:
IrinaVladis [17]3 years ago
5 0

Answer:

Q9. The independent variable in this experiment is the fertilizer. It is independent because she manipulating the variable to compare the growth.

Q10. The dependent variable in this experiment is the amount of growth of the corn. It is this because the growth depends on what the scientist did on the corn.

Q11. The variable controlled in this experiment is the amount of sun and water. These two variables never change so this is why it is the control.

Explanation:

QveST [7]3 years ago
5 0

Answer:

Explanation:

independent - 5 months dependent - sun controlled - water and the fertilizer

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Property of Volume changes.

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A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
Mashcka [7]

Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

(a)  Relation between K_{a} and pK_{a} is as follows.

                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to form HNO_{2}

Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = pK_{a} + log \frac{[conjugate base]}{[acid]}

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

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Answer:

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Explanation:

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The pH of a solution is 3.81. What is the OH concentration in the solution?​
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Answer:

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Explanation:

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pH of the solution: 3.81

Step 2: Calculate the pOH of the solution

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Step 3: Calculate the concentration of OH⁻ ions

We will use the definition of pOH.

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